CBSE Class 8 Maths — Squares and Square Roots

Chapter Overview & Weightage

Squares and Square Roots is one of the most fundamental chapters in Class 8 Maths — the concepts here support everything from Pythagoras theorem to quadratic equations in higher classes. In the CBSE Class 8 annual exam, this chapter typically contributes 8–10 marks within the Number System unit.

Key Concepts You Must Know

Perfect square: A natural number that is the square of another natural number. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100… are perfect squares.

Properties of perfect squares (exam favourites):

• Perfect squares always end in 0, 1, 4, 5, 6, or 9. They never end in 2, 3, 7, or 8.
• A perfect square has an odd number of factors.
• The square of an even number is even; square of an odd number is odd.
• Between consecutive squares $n^2$ and $(n+1)^2$, there are exactly $2n$ non-square numbers.
• Sum of first $n$ odd numbers = $n^2$ (useful for finding perfect squares quickly).

Pythagorean triplets: Three numbers $(a, b, c)$ where $a^2 + b^2 = c^2$. For any natural number $m > 1$: $(2m,\, m^2-1,\, m^2+1)$ is a Pythagorean triplet. Common triplets: (3,4,5), (5,12,13), (8,15,17), (7,24,25).

Square root: The inverse of squaring. $\sqrt{a^2} = a$ (for $a \geq 0$). Two methods: (1) Prime factorisation, (2) Long division.

Important Formulas

Solved Previous Year Questions

PYQ 1 — Identify and find smallest multiplier (2-mark type)

Q: Find the smallest number by which 180 must be multiplied to make it a perfect square. Also find the square root of the result.

Solution: Prime factorisation: $180 = 2^2 \times 3^2 \times 5$.

The factor 5 has no pair. So we multiply by 5: $180 \times 5 = 900$.

$900 = 2^2 \times 3^2 \times 5^2$

$\sqrt{900} = 2 \times 3 \times 5 = 30$.

The smallest multiplier is 5; square root of 900 is 30.

PYQ 2 — Long division method (3-mark type)

Q: Find $\sqrt{1521}$ using the long division method.

Solution: Group digits from right: 15 | 21.

• Largest square ≤ 15 is $3^2 = 9$. Write 3 as first digit of root.
• Remainder: $15 - 9 = 6$. Bring down 21: new dividend = 621.
• Double the quotient: $2 \times 3 = 6$. Find digit $d$ such that $6d \times d \leq 621$.
• Try $d = 9$: $69 \times 9 = 621$. Exactly 621. ✓

$\sqrt{1521} = \mathbf{39}$. Verify: $39^2 = 1521$ ✓.

PYQ 3 — Pythagorean triplet (1-mark type)

Q: Find a Pythagorean triplet whose smallest member is 14.

Solution: Use the formula: smallest member = $2m$. So $2m = 14 \Rightarrow m = 7$.

Triplet: $(2m, m^2-1, m^2+1) = (14, 48, 50)$.

Check: $14^2 + 48^2 = 196 + 2304 = 2500 = 50^2$ ✓.

Difficulty Distribution

Expert Strategy

The long division method for square roots is the most feared calculation in this chapter — but it has a perfectly repeatable algorithm. Practise it 10 times with different numbers and it becomes mechanical. The key rule: always group digits in pairs from the decimal point — left for whole numbers, right for decimals.

For “find the smallest number to multiply/divide to get a perfect square” — always start with prime factorisation. Then identify the “lone” factors (those without a pair). Multiply by them to complete pairs; divide by them to eliminate unpaired factors.

Common Traps